why do S1 and S2 behave differently with respect to apostrophe location?
S1="1/282/03/10"
S2="4/107/03/10"
R1="".join({"N\'" ,S1,"\'" })
R2="".join({"N\'" ,S2,"\'" })
Outputs:
"N''1/282/03/10" //R1
"N'4/107/03/10'" //R2
forward slash in string literal behaves unexpectedly
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2 Answers
That's because you are using set.
setis a collection which is unordered and unindexed.
So you have to use the following code to get the desired results,
S1 = "1/282/03/10"
S2 = "4/107/03/10"
R1 = "".join(["N\'", S1, "\'"])
R2 = "".join(["N\'", S2, "\'"])
print(R1)
print(R2)
Output:
N'1/282/03/10'
N'4/107/03/10'
The reason is you wrap "\'",S1,"'\" in a set.
set is an unordered collection.
To avoid the above issue use list or tuple.
R1 = "".join(["N\'", S1, "\'"])
R2 = "".join(["N\'", S2, "\'"])
#tuple equivalent is
R1 = "".join(("N\'", S1, "\'"))
R2 = "".join(("N\'", S2, "\'"))
You can use .format here you don't even need to use list or tuple.
R1="N'{}'".format(S1)
R2="N'{}'".format(S2)
You can also use this.
R1 = "N\'" + S1 + "\'"
R2 = "N\'" + S2 + "\'"
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